Say the definition of a propagator in quantum field theory is:

$$G_F(x,y)=\int \phi(x)\phi(y) e^{i S[\phi] } D\phi$$

where $S$ is the action. Why do we integrate the Lagrangian density from $t=-\infty$ to $t=+\infty$ instead of from $x_0$ to $y_0$?


$$S[\phi] = \int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty} {\cal{L}}(x,y,z,t)dxdydzdt$$

where $\cal{L}$ is the Lagrangian density. Surely we are interested only in the section between $t=x_0$ and $t=y_0$?


2 Answers 2


Short answer (assuming proper time-ordering of $x$ and $y$):

$$ \langle \Omega| \phi(x) \phi(y) | \Omega\rangle = \frac{ \langle 0 | U(-\infty, x^0) \phi_I(x) U(x^0, y^0) \phi_I(y) U(y^0, +\infty) |0 \rangle }{\langle 0 | U(-\infty, \infty) | 0 \rangle } $$

These infinities has to be in the action integral too.

Now there is a rather deep reason for having time going to infinity in the path integral (the credit for this goes to Weinberg and his awesome book, as usual). The propagators usually has funny $i \epsilon$ in denominators, like:

$$ \frac{1}{q^2 + m^2 - i \epsilon} $$

Which makes possible integration over $q$ and helps select proper pole with Residue theorem.

Although it is usually just assumed and omitted in calculation, this $i \epsilon$ comes from the phases of field at time $\pm \infty$:

$$ \langle \Omega,out| \phi(\infty) \rangle \langle \phi(-\infty) | \Omega, in \rangle \propto e^{\epsilon \, \times \, \cdots \text{pairs of fields} } $$

for some infinitesimal $\epsilon$.

(Note that in interacting theory it is possible to calculate this product of projections only at $t = \pm \infty$, because it is assumed that the theory is free at this time.)

Given that this product is proportional to the exponential form - the product of $\epsilon$ and pairs of fields will be merged with the action! This is the reason to have $\epsilon$ in the propagator.

(You may find more details in Chapter 9 of Weinberg's QFT Volume 1)

  • $\begingroup$ Thanks, its a bit too concise for me! The gist seems to be that you have to start with the vacuum in the distance past and future. $\endgroup$
    – user84158
    Sep 30, 2017 at 3:10
  • $\begingroup$ Elaborated on this. Hopefully you'll enjoy the answer. $\endgroup$
    – Darkseid
    Sep 30, 2017 at 3:16
  • $\begingroup$ @zooby That's right - the physical reason for this is that the vacuum state of an interacting theory is not the same as the vacuum state of the corresponding free theory. They are close in the distant past and future, when the interacting "particles" are well-separated and the interaction terms are small, but they aren't identical. $\endgroup$
    – J. Murray
    Sep 30, 2017 at 3:21
  • $\begingroup$ In the context of Feynman diagrams, the difference between the interacting vacuum and the free vacuum is accounted for by discarding certain diagrams (e.g. the un-amputated ones) and adding various normalization factors - in other words, the LSZ reduction formula. $\endgroup$
    – J. Murray
    Sep 30, 2017 at 3:28
  • $\begingroup$ So is $G(x,y)$ not really the amplitude for a particle to get from x to y but the amplitude for a particle to come in from the distant past pass through x and y and leave to the distant future? $\endgroup$
    – user84158
    Sep 30, 2017 at 14:16

This is a common theme in physics: If you have a separation of characteristic scales in the problem, physical effects associated with sub-dominant scales can be ignored.

E.g. in $S$-matrix theory, it is often assumed for simplicity, that the interactions can be neglected near the initial and final times, $t=t_i$ and $t=t_f$, and that the time interval $[t_i,t_f]$ is much bigger than all the characteristic time scales of the interactions.

In other words, if we are not interested in studying temporal boundary effects, we might as well send $t_i\to -\infty$ and $t_f\to \infty$ to simplify the problem.

  • 1
    $\begingroup$ And also (to be completely honest) this is the only regime where we know how to compute the $S$ operator perturbatively for an interacting QFT, right? $\endgroup$ Sep 30, 2017 at 8:37

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